Skip to Content
Stochastic Process

Poisson Process

2020-01-31Original-language archivelegacy assets may be incomplete

Poisson Process

Counting Process

  1. N(t)0N(t)\geq 0
  2. N(t)NN(t)\in\mathbb{N}
  3. s<t,N(s)N(t)s<t,N(s)\leq N(t)
  4. s<t,N(t)N(s)s<t,N(t)-N(s) equals the number of events that have occurred in the interval (s,t](s,t]

Definition 1 of Poisson Process

  1. N(0)=0N(0)=0

  2. independent increments

  3. s,t0\forall s,t\geq 0

    P(N(t+s)N(s)=n)=eλt(λt)nn!,n=0,1P(N(t+s)-N(s)=n)=e^{-\lambda t}\frac{(\lambda t)^n}{n!},n=0,1

Definition 2 of Poisson Process

  1. N(0)=0N(0)=0
  2. The process has stationary and independent increments
  3. P(N(t+h)N(t)=1)=λh+o(h)P(N(t+h)-N(t)=1)=\lambda h+o(h)
  4. P(N(t+h)N(t)2)=o(h)P(N(t+h)-N(t)\geq 2)=o(h)

Sequence of interarrival times

  • {Xn,n1}\{X_n,n\geq 1\}: XnX_n denotes the time of the nnth event

    P(X1>t)=P(N(t)=0)=eλtP(X_1>t)=P(N(t)=0)=e^{-\lambda t}

  • XnX_n are independent identically distributed exponential random variables having mean 1λ\frac{1}{\lambda}

Arrival time of nnth event

  • Sn=i=1nXi,n1S_n=\sum_{i=1}^nX_i,n\geq 1: Gamma Distribution

  • order statistics Y(i)Y_{(i)} corresponding to YiY_i: Y(i)Y_{(i)} is the kkth minimum of YiY_{i}

    f(y1,y2,,yn)=n!i=1nf(yi)f(y_1,y_2,\cdots,y_n)=n!\prod_{i=1}^nf(y_i)

  • YiY_i are uniformly distributed over (0,t)(0,t), then the joint density funciton of order statistics is

    f(y1,y2,,yn)=n!tnf(y_1,y_2,\cdots,y_n)=\frac{n!}{t^n}

  • Given that N(t)=nN(t)=n, the nn arrival times S1,,SnS_1,\cdots,S_n have the same distribution as the order statistics corresponding to nn independent random variables uniformly distributed on the interval (0,t)(0,t)

Two Type Poisson random variables

  • event occurs at time ss, then, indepentently of all else, it is classified as being a type-I event with probability P(s)P(s) and a type-II event with probability 1P(s)1-P(s)
  • N1(t)N_1(t) and N2(t)N_2(t) are independent Poisson random variables having respective means λtp\lambda tp and λt(1p)\lambda t(1-p)

p=1t0tP(s)dsp=\frac{1}{t}\int^t_0P(s)ds

M/G/1 Busy Period

  • M/G/1 queueing system
    • customers arrive in accordance with a Poisson process with rate λ\lambda
    • upon arrival, either enter service if the server is free or else join the queue
    • successive service times are independent and identically distributed according to GG: YiY_i denotes the sequence of service times
    • busy period begins: an arrival finds the server free
    • busy period ends: no longer any customers in the system
  • busy period last a time tt and consits of nn services (Probability B(t,n)B(t,n)) iff
    • SkY1+YkS_k\leq Y_1+\cdots Y_k
    • Y1+Yn=tY_1+\cdots Y_n=t
    • There are n1n-1 arrivals in (0,t)(0,t)

B(t,n)=0teλt(λt)n1n!dGn(t)B(t,n)=\int_0^te^{-\lambda t}\frac{(\lambda t)^{n-1}}{n!}dG_n(t)

B(t)=n=1B(t,n)B(t)=\sum_{n=1}^\infty B(t,n)